Outline. Probability. Math 143. Department of Mathematics and Statistics Calvin College. Spring 2010

Transcription

1 Outline Math 143 Department of Mathematics and Statistics Calvin College Spring 2010

2 Outline Outline 1 Review Basics Random Variables Mean, Variance and Standard Deviation of Random Variables 2 More

3 Review More Quick Review of Basics Random Variables Mean, Variance and Standard Deviation of Random Variables measures long-term relative frequency. Empirical: P(E) = # of times E happens # of times random procedure is repeated Theoretical: Based on some basic probability rules (always true) assumptions about the particular situation at hand mathematical reasoning Discrete: Assign probablity to each outcome (table works well) Continuous: = area under density curve

4 Basic Rules Review More Basics Random Variables Mean, Variance and Standard Deviation of Random Variables 1 0 P(E) 1 for any event E. 2 P(S) = 1, where S is the sample space. 3 P(A or B) = P(A) + P(B) provided A and B are mutually exclusive. These imply 4 a simple rule for equally likely outcomes: P(E) = # of outcomes in E # of outcomes in S 5 P(E c ) = 1 P(E) [so P(E) = 1 P(E c ),too]

5 Random Variables Review More Basics Random Variables Mean, Variance and Standard Deviation of Random Variables A random variable is a numerical result of a random process. A distribution of a random variable tells us the possible values of a random variable (what values?), and the probability of having those values (with what frequency/proportion?). We use random variables to help us quantify the results of random processes especially random sampling. Random variables are generally denoted using capital letters, like X, Y, Z.

6 Review More Examples of random variables Basics Random Variables Mean, Variance and Standard Deviation of Random Variables 1 Deal a poker hand. Let X = the number of face cards. X assigns to 3, 7, 10, Q, K the value 2. 2 Roll two six-sided dice. Let Y = the sum of the two numbers rolled. Y assigns to a roll of {2, 4} the value 6. 3 Let H = the difference between the heights of a randomly selected couple (height of husband height of wife, measured in inches). If my wife and I were the randomly selected couple, then the value of H would be A family plans to have four children. Let Y be the number of boys. 5 Randomly sample 100 people and ask them a yes/no question. Let X be the number of people in the sample who answer yes. There are two kinds of random variables: discrete and continuous.

7 Review More Discrete Random Variables Basics Random Variables Mean, Variance and Standard Deviation of Random Variables If X is discrete (takes on values from a list possibilities), we can describe the distribution of X by making a table that lists each possible value of X and the probability that it occurs. Value of X x 1 x 2 x 3... x n p 1 p 2 p 3... p n Of course, each probability (p i ) must be between 0 and 1 (inclusive), and the sum of all the probabilities ( p i ) must equal 1.

8 Review More Flipping Three Fair Coins Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Example (Flipping 3 fair coins and counting heads) Flip 3 fair coins. Let X = the number of heads. Find the probability distribution of X. Determine P(X 2). Determine P(X is even).

9 Review More Flipping Three Fair Coins Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Example (Flipping 3 fair coins and counting heads) Flip 3 fair coins. Let X = the number of heads. Find the probability distribution of X. Determine P(X 2). Determine P(X is even). Value of X

10 Review More Flipping Three Fair Coins Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Example (Flipping 3 fair coins and counting heads) Flip 3 fair coins. Let X = the number of heads. Find the probability distribution of X. Determine P(X 2). Determine P(X is even). Value of X /8 3/8 3/8 1/8

11 Review More Flipping Three Fair Coins Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Example (Flipping 3 fair coins and counting heads) Flip 3 fair coins. Let X = the number of heads. Find the probability distribution of X. Determine P(X 2). Determine P(X is even). Value of X /8 3/8 3/8 1/8 P(X 2) = 3/8 + 1/8 = 1/2

12 Review More Flipping Three Fair Coins Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Example (Flipping 3 fair coins and counting heads) Flip 3 fair coins. Let X = the number of heads. Find the probability distribution of X. Determine P(X 2). Determine P(X is even). Value of X /8 3/8 3/8 1/8 P(X 2) = 3/8 + 1/8 = 1/2 P(X is even) = 1/8 + 3/8 = 1/2

13 Histograms Review More Basics Random Variables Mean, Variance and Standard Deviation of Random Variables We can also describe this probability distribution graphically, using a probability histogram Value of X /8 3/8 3/8 1/8

14 Histograms Review More Basics Random Variables Mean, Variance and Standard Deviation of Random Variables We can also describe this probability distribution graphically, using a probability histogram Value of X /8 3/8 3/8 1/

15 Personality Inventory Review More Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Example Sometimes we base a model for a random variable on data collected from a sample. For example, an industrial psychologist administered a personality inventory test for passive-aggressive traits to 150 employees. Individuals were rated on a scale from 1 to 5 (passive to aggressive). The results are as follows: Score Frequency

16 Personal Inventory Review More Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Let T be rating of a random employee. Assuming the 150 tested employees are typical of all the employees, we can construct the probability distribution table for the random variable T. Value of T Count Now compute the following P(T > 4) =

17 Personal Inventory Review More Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Let T be rating of a random employee. Assuming the 150 tested employees are typical of all the employees, we can construct the probability distribution table for the random variable T. Value of T Count Now compute the following P(T > 4) =.14 P(T 4) =

18 Personal Inventory Review More Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Let T be rating of a random employee. Assuming the 150 tested employees are typical of all the employees, we can construct the probability distribution table for the random variable T. Value of T Count Now compute the following P(T > 4) =.14 P(T 4) = =.34 P(T 2) =

19 Personal Inventory Review More Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Let T be rating of a random employee. Assuming the 150 tested employees are typical of all the employees, we can construct the probability distribution table for the random variable T. Value of T Count Now compute the following P(T > 4) =.14 P(T 4) = =.34 P(T 2) = =.38 P(3 T 4) =

20 Personal Inventory Review More Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Let T be rating of a random employee. Assuming the 150 tested employees are typical of all the employees, we can construct the probability distribution table for the random variable T. Value of T Count Now compute the following P(T > 4) =.14 P(T 4) = =.34 P(T 2) = =.38 P(3 T 4) = =.48

21 Personal Inventory Review More Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Let T be rating of a random employee. Assuming the 150 tested employees are typical of all the employees, we can construct the probability distribution table for the random variable T. Value of T Count Now compute the following P(T > 4) =.14 P(T 4) = =.34 P(T 2) = =.38 P(3 T 4) = =.48 Suppose this test were given to 500 employees. Based on the above probability distribution, about how many would we expect to be extremely aggressive (score of 5)?

22 Review More Continuous Random Variables Basics Random Variables Mean, Variance and Standard Deviation of Random Variables A continuous random variable can take all values in an interval of numbers. Therefore the probability distribution of a continuous random variable can t be described by making a table of values and probabilities. Instead, the probability distribution of a continuous random variable X is described by a density curve. Definition The probability of any event is the area under the density curve and above the values of X that make up the event. Just like we have been drawing for normal distributions Use calculus to compute (or estimate) the area Important statistical distributions are built into software, and included in tables like the one for normal distributions

23 Review More Probabilities from Density Curves Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Example (Simple Density Curve) Below is a simple density curve for a random variable X. 1 2 Based on the density curve, we can compute things like P(X > 1) P(.5 X 1.5) P(X = 1)

24 Review More Notation for Normal Distributions Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Recall that the normal distributions are a family of continuous probability distributions with distinctive bell-shaped density curves. If X is normally distributed with mean µ and standard deviation σ, we will write X N(µ, σ). If X is approximately normally distributed, we will write X N(µ, σ). Example (Using this notation) Suppose Z N(0, 1), then P( 1 Z 1) =

25 Review More Notation for Normal Distributions Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Recall that the normal distributions are a family of continuous probability distributions with distinctive bell-shaped density curves. If X is normally distributed with mean µ and standard deviation σ, we will write X N(µ, σ). If X is approximately normally distributed, we will write X N(µ, σ). Example (Using this notation) Suppose Z N(0, 1), then P( 1 Z 1) = 0.68 (approx.)

26 Review More Notation for Normal Distributions Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Recall that the normal distributions are a family of continuous probability distributions with distinctive bell-shaped density curves. If X is normally distributed with mean µ and standard deviation σ, we will write X N(µ, σ). If X is approximately normally distributed, we will write X N(µ, σ). Example (Using this notation) Suppose Z N(0, 1), then P( 1 Z 1) = 0.68 (approx.) But what does it mean to take the mean or standard deviation of a random variable?

27 Review More Computing the Mean from Data Basics Random Variables Mean, Variance and Standard Deviation of Random Variables When we want to compute the mean of a list of numbers {x 1, x 2,... x n }, we simply add the numbers and divide by how many there were mean = x = xi n = x 1 + x 2 + x n n

28 Review More Computing the Mean from Data Basics Random Variables Mean, Variance and Standard Deviation of Random Variables When we want to compute the mean of a list of numbers {x 1, x 2,... x n }, we simply add the numbers and divide by how many there were mean = x = xi This can be written another way: n = x 1 + x 2 + x n n mean = x = x 1 n + x 2 n + xn n = x 1 1 n + x 2 1 n + x n 1 n = x i 1 n

29 Review More Computing the Mean from Data Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Written this last way, we can figure out how to compute the mean (also called expected value) of a discrete random variable. mean = x = x 1 1 n + x 2 1 n + x n 1 n = x i 1 n Imagine choosing one of our numbers {x 1, x 2,... x n } at random, each equally likely. Then 1 n is the probability of choosing a particular number in our list. So we see that Key Idea the mean is the sum of values times probabilities usually call this a weighted sum

30 Review More The Mean of a Discrete Random Variable Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Suppose X is a discrete random variable and the distribution of X is given by Value of X x 1 x 2 x 3... x n p 1 p 2 p 3... p n Then to find the mean of X, multiply each possible value by its probability, then add all the products: Definition (Mean of Discrete Random Variable) mean of X = µ X = x i p i

31 Review More Flipping Three Fair Coins Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Example (Flipping 3 fair coins and counting heads) Let X be the random variable described by What is µ X? Value of X /8 3/8 3/8 1/8

32 Review More Flipping Three Fair Coins Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Example (Flipping 3 fair coins and counting heads) Let X be the random variable described by What is µ X? Value of X /8 3/8 3/8 1/ = 1.5

33 Review More Variance and Standard Deviation Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Definition Suppose X is a discrete random variable and the distribution of X is given by Value of X x 1 x 2 x 3... x n p 1 p 2 p 3... p n Then to find the variance of X, multiply the squared difference from the mean for each possible value by its probability, then add all the products: variance of X = σ 2 X = (x i µ X ) 2 p i The standard deviation is the square root of the variance: σ X = σ 2.

34 Review More Flipping Three Fair Coins Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Example (Flipping 3 fair coins and counting heads) Let X be the random variable described by Value of X /8 3/8 3/8 1/8 What is σ X? (Recall that µ X = 1.5.)

35 Review More Flipping Three Fair Coins Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Example (Flipping 3 fair coins and counting heads) Let X be the random variable described by Value of X /8 3/8 3/8 1/8 What is σ X? (Recall that µ X = 1.5.) σx 2 = (0 1.5) (1 1.5) (2 1.5) (3 1.5)2 1 8 = = 0.75 σ X =.75 =

36 Review More Basics Random Variables Mean, Variance and Standard Deviation of Random Variables Mean and Variance for Continuous Random Variables The means and standard deviations of continuous random variables are computed in a related way, but require more advanced mathematics (integral calculus) to express precisely. Basically, sums are replaced by integrals: Definition (You are NOT required to know this.) µ X = xp(x) dx (σ X ) 2 = (x µ x ) 2 p(x) dx

37 Adding Another Rule 1 0 P(E) 1 for any event E. 2 P(S) = 1, where S is the sample space. 3 P(E c ) = 1 P(E) [so P(E) = 1 P(E c ),too] 4 P(A or B) = P(A) + P(B) provided A and B are mutually exclusive. 5 P(A and B) = P(A) P(B) provided A and B are independent. Two events are independent if knowing whether one of them occurs or not does not change the probability of the other.

38 Adding Another Rule 1 0 P(E) 1 for any event E. 2 P(S) = 1, where S is the sample space. 3 P(E c ) = 1 P(E) [so P(E) = 1 P(E c ),too] 4 P(A or B) = P(A) + P(B) provided A and B are mutually exclusive. 5 P(A and B) = P(A) P(B) provided A and B are independent. Two events are independent if knowing whether one of them occurs or not does not change the probability of the other. Example (Independent Events) Toss two fair coins. Getting a head on the second toss is independent of getting a head on the first toss, since the probability is still 1/2 for getting a head on the second toss, even if we know the first toss was a head (or a tail).

39 Yahtzee Review More A standard 6-sided die has six sides numbered 1, 2, 3, 4, 5, and 6. Each number is equally likely to be rolled (assuming a fair die). Example (Yahtzee ) In the game of Yahtzee, 5 six-sided dice are rolled. If all 5 dice have the same number, the roll is called a Yahtzee. What is the probability of rolling 5 dice and getting a Yahtzee?

40 Review More An Easier Related Problem Let s try something easier to start: Example (Easier) If we roll two dice, what is the probability that the numbers rolled match? (That is, what is the probability of rolling doubles?)

41 Review More An Easier Related Problem Let s try something easier to start: Example (Easier) If we roll two dice, what is the probability that the numbers rolled match? (That is, what is the probability of rolling doubles?) Method 1: There are 36 possible rolls (for each of 6 possibilities on die 1, there are 6 possibilities on die 2). Of these only 6 are doubles. So the probability is 6 36 = 1 6.

42 Review More An Easier Related Problem Let s try something easier to start: Example (Easier) If we roll two dice, what is the probability that the numbers rolled match? (That is, what is the probability of rolling doubles?) Method 1: There are 36 possible rolls (for each of 6 possibilities on die 1, there are 6 possibilities on die 2). Of these only 6 are doubles. So the probability is 6 36 = 1 6. Method 2: Change the question to After we roll the first die, what is the probability that the second one will match it? which has the same answer. There are 6 equally likely outcomes of the second die, only one matches, so the answer is 1/6, same as we got using method 1.

43 Yahtzee Review More Example (original Yahtzee ) In the game of Yahtzee, 5 six-sided dice are rolled. If all 5 dice have the same number, the roll is called a Yahtzee. What is the probability of rolling 5 dice and getting a Yahtzee?

44 Yahtzee Review More Example (original Yahtzee ) In the game of Yahtzee, 5 six-sided dice are rolled. If all 5 dice have the same number, the roll is called a Yahtzee. What is the probability of rolling 5 dice and getting a Yahtzee? Method 1: Now there are = 6 5 = 7776 possible outcomes. Again only 6 are Yahtzees, so the probability is = =

45 Yahtzee Review More Example (original Yahtzee ) In the game of Yahtzee, 5 six-sided dice are rolled. If all 5 dice have the same number, the roll is called a Yahtzee. What is the probability of rolling 5 dice and getting a Yahtzee? Method 1: Now there are = 6 5 = 7776 possible outcomes. Again only 6 are Yahtzees, so the probability is = = Method 2: After the first die is rolled, the probability that each subsequent die matches it is 1/6. These events are independent, so we can multiply to get the probability that they all happen at once: = 1 =

46 Mendel s Peas Review More The peas from Mendel s genetics experiments could be either yellow or green. In this simple situation, color is determined by a single gene. Each organism inherits two of these genes (one from each parent), which can be one of two types (Y,y). A pea plant produces yellow peas unless BOTH genes are type y. As a first step, Mendel bred pure line pea plants of type YY or yy. The offspring of these plants are very predictable: YY YY : all offspring are YY (and therefore yellow) yy yy: all offspring are yy (and therefore green) YY yy: all offspring are Yy (and therefore yellow)

47 Mendel s Peas Review More More interestingly, consider Yy Yy: P(green) = P(yy) =

48 Mendel s Peas Review More More interestingly, consider Yy Yy: P(green) = P(yy) = P(y from 1st Yy and y from 2nd Yy) = P(y from 1st) P(y from 2nd) = 1/2 1/2 = 1/4

49 Mendel s Peas Review More More interestingly, consider Yy Yy: P(green) = P(yy) = P(y from 1st Yy and y from 2nd Yy) = P(y from 1st) P(y from 2nd) = 1/2 1/2 = 1/4 P(yellow) =

50 Mendel s Peas Review More More interestingly, consider Yy Yy: P(green) = P(yy) = P(y from 1st Yy and y from 2nd Yy) = P(y from 1st) P(y from 2nd) = 1/2 1/2 = 1/4 P(yellow) = 1 P(green) = 3/4.

51 Mendel s Peas Review More More interestingly, consider Yy Yy: P(green) = P(yy) = P(y from 1st Yy and y from 2nd Yy) = P(y from 1st) P(y from 2nd) = 1/2 1/2 = 1/4 P(yellow) = 1 P(green) = 3/4. P(heterozygote) = P(Yy or yy )

52 Mendel s Peas Review More More interestingly, consider Yy Yy: P(green) = P(yy) = P(y from 1st Yy and y from 2nd Yy) = P(y from 1st) P(y from 2nd) = 1/2 1/2 = 1/4 P(yellow) = 1 P(green) = 3/4. P(heterozygote) = P(Yy or yy ) = 1/4 + 1/4 = 1/2

53 Mendel s Peas Review More More interestingly, consider Yy Yy: P(green) = P(yy) = P(y from 1st Yy and y from 2nd Yy) = P(y from 1st) P(y from 2nd) = 1/2 1/2 = 1/4 P(yellow) = 1 P(green) = 3/4. P(heterozygote) = P(Yy or yy ) = 1/4 + 1/4 = 1/2 Notice where we are repeatedly using the assumptions: Which gene is inherited from a parent is equally likely The genes inherited from the parents are independent of each other We can look at how well the data fit the theoretical probabilities to test the assumptions. (Mendel s data fit very well.)

54 More Peas Review More Mendel actually looked at two traits: color (yellow,green) and shape (round, wrinkled). According to his model, yellow (Y) and round were dominant (R). If these traits are inherited independently, then we can make predictions about a YyRr YyRr cross: So P(yellow) = 3 4 P(green) = 1 4 P(round) = 3 4 P(wrinkled) = 1 4 P(yellow and round) =

55 More Peas Review More Mendel actually looked at two traits: color (yellow,green) and shape (round, wrinkled). According to his model, yellow (Y) and round were dominant (R). If these traits are inherited independently, then we can make predictions about a YyRr YyRr cross: So P(yellow) = 3 4 P(green) = 1 4 P(round) = 3 4 P(wrinkled) = 1 4 P(yellow and round) = = P(yellow and wrinkled) =

56 More Peas Review More Mendel actually looked at two traits: color (yellow,green) and shape (round, wrinkled). According to his model, yellow (Y) and round were dominant (R). If these traits are inherited independently, then we can make predictions about a YyRr YyRr cross: So P(yellow) = 3 4 P(green) = 1 4 P(round) = 3 4 P(wrinkled) = 1 4 P(yellow and round) = = P(yellow and wrinkled) = = P(green and round) =

57 More Peas Review More Mendel actually looked at two traits: color (yellow,green) and shape (round, wrinkled). According to his model, yellow (Y) and round were dominant (R). If these traits are inherited independently, then we can make predictions about a YyRr YyRr cross: So P(yellow) = 3 4 P(green) = 1 4 P(round) = 3 4 P(wrinkled) = 1 4 P(yellow and round) = = P(yellow and wrinkled) = = P(green and round) = = P(green and wrinkled) =

58 More Peas Review More Mendel actually looked at two traits: color (yellow,green) and shape (round, wrinkled). According to his model, yellow (Y) and round were dominant (R). If these traits are inherited independently, then we can make predictions about a YyRr YyRr cross: So P(yellow) = 3 4 P(green) = 1 4 P(round) = 3 4 P(wrinkled) = 1 4 P(yellow and round) = = P(yellow and wrinkled) = = P(green and round) = = P(green and wrinkled) = =

59 More Peas Review More Mendel actually looked at two traits: color (yellow,green) and shape (round, wrinkled). According to his model, yellow (Y) and round were dominant (R). If these traits are inherited independently, then we can make predictions about a YyRr YyRr cross: So P(yellow) = 3 4 P(green) = 1 4 P(round) = 3 4 P(wrinkled) = 1 4 P(yellow and round) = = P(yellow and wrinkled) = = P(green and round) = = P(green and wrinkled) = =

60 Review More Flipping Three Unfair Coins Example (Flipping 3 unfair coins and counting heads) Flip 3 coins that result in heads 3/4 of the time. Let H = the number of heads. Find the probability distribution of H. Determine P(H 2). Determine P(H is even).

61 Review More Flipping Three Unfair Coins Example (Flipping 3 unfair coins and counting heads) Flip 3 coins that result in heads 3/4 of the time. Let H = the number of heads. Find the probability distribution of H. Determine P(H 2). Determine P(H is even). Value of H

62 Review More Flipping Three Unfair Coins Example (Flipping 3 unfair coins and counting heads) Flip 3 coins that result in heads 3/4 of the time. Let H = the number of heads. Find the probability distribution of H. Determine P(H 2). Determine P(H is even). Value of H /64 9/64 27/64 27/64

63 Review More Flipping Three Unfair Coins Example (Flipping 3 unfair coins and counting heads) Flip 3 coins that result in heads 3/4 of the time. Let H = the number of heads. Find the probability distribution of H. Determine P(H 2). Determine P(H is even). Value of H /64 9/64 27/64 27/64 P(H = 1) = = = 9 64

64 Review More Flipping Three Unfair Coins Example (Flipping 3 unfair coins and counting heads) Flip 3 coins that result in heads 3/4 of the time. Let H = the number of heads. Find the probability distribution of H. Determine P(H 2). Determine P(H is even). Value of H /64 9/64 27/64 27/64 P(H = 1) = P(H 2) = 27/ /64 = 54/ = = 9 64

65 Review More Flipping Three Unfair Coins Example (Flipping 3 unfair coins and counting heads) Flip 3 coins that result in heads 3/4 of the time. Let H = the number of heads. Find the probability distribution of H. Determine P(H 2). Determine P(H is even). Value of H /64 9/64 27/64 27/64 P(H = 1) = P(H 2) = 27/ /64 = 54/64 P(H is even) = 1/ /64 = 28/ = = 9 64

66 Review More What s to Come: Generalized Rules We still need to answer the following questions: What is P(A or B) when A and B are NOT mutually exclusive? What is P(A and B) when A and B are NOT independent?

67 General Addition Rule Review More We can replace our old rule #4 with a more general version. 4 For any events A and B, P(A or B) =

68 General Addition Rule Review More We can replace our old rule #4 with a more general version. 4 For any events A and B, P(A or B) = P(A) + P(B) P(A and B).

69 General Addition Rule Review More We can replace our old rule #4 with a more general version. 4 For any events A and B,. Example (Rolling Dice) P(A or B) = P(A) + P(B) P(A and B) If we roll two fair 6-sided dice, what is the probability of getting at least one 5?

70 General Addition Rule Review More We can replace our old rule #4 with a more general version. 4 For any events A and B,. Example (Rolling Dice) P(A or B) = P(A) + P(B) P(A and B) If we roll two fair 6-sided dice, what is the probability of getting at least one 5? P(first is 5) = 1/6; P(second is 5) = 1/6 (equally likely outcomes) P(both are 5) = (1/6) (1/6) = 1/36 (independent events) So P(at least one is 5) = = 11 36

71 Conditional Review More Consider the probability chart below which shows the probabilities for sex and favorite color in Mr. Ortez s class. Red Yellow Blue boy girl If a student is selected at random from Mr. Ortez s class: What is the probability of selecting a boy? What is the probability that the student s favorite color is blue, given that the student is a girl? What is the probability that the student is a girl, given that the favorite color is blue?

72 Conditional Review More Motivated by the preceeding example, we make the following definition: Definition (Conditional ) P(B A) = P(A and B) P(A)

73 Conditional Review More Motivated by the preceeding example, we make the following definition: Definition (Conditional ) P(B A) = P(A and B) P(A) Note that the following statements are equivalent: A and B are independent P(A and B) = P(A) P(B) P(B A) = P(A) P(B) P(A) = P(B) So the probability of B is unchanged by knowledge of whether A occurs.

74 General Product Rule Review More Definition (Conditional ) P(B A) = If we rewrite this definition, we get General Product Rule For any events A and B, P(A and B) = P(A and B) P(A)

75 General Product Rule Review More Definition (Conditional ) P(B A) = If we rewrite this definition, we get General Product Rule For any events A and B, P(A and B) P(A) P(A and B) = P(A) P(B A)

76 Review More General Product Rule (Example) Example (Same Sex Selection) Suppose a class has 10 men and 10 women. If three students are selected at random, what is the probability that all three are the same sex?

77 Review More General Product Rule (Example) Example (Same Sex Selection) Suppose a class has 10 men and 10 women. If three students are selected at random, what is the probability that all three are the same sex? Let S = second person is the same sex as the first. Let T = third person is the same sex as the first. P(S and T ) = P(S) P(T S) =

78 Review More General Product Rule (Example) Example (Same Sex Selection) Suppose a class has 10 men and 10 women. If three students are selected at random, what is the probability that all three are the same sex? Let S = second person is the same sex as the first. Let T = third person is the same sex as the first. P(S and T ) = P(S) P(T S) = = 4 19 Note: this is less than the probability of getting the same value on a coin if we flip it three times because it gets harder and harder to get someone of the same sex as we use them up. (The coin probability is 1 4.)

79 From the Little Survey Review More Two Versions of a Question Social science researchers have conducted extensive empirical studies and concluded that the expression absence makes the heart grow fonder out of sight out of mind is generally true. Do you find this result surprising or not surprising? Results (previous class): A absence A c sight surprised (S) 6 5 not surprised (S c ) Q. Is being surprised independent of the form of the question asked?

80 From the Little Survey Review More Results (from a previous class): A absence A c sight surprised (S) 6 5 not surprised (S c ) Is being surprised independent of the form of the question asked?

81 From the Little Survey Review More Results (from a previous class): A absence A c sight surprised (S) 6 5 not surprised (S c ) Is being surprised independent of the form of the question asked? = P(A) = P(S) = P(S and A) = 6 91 = P(A) P(S) = P(S A) = 6/91 50/91 = =

82 From the Little Survey Review More Results (from a previous class): A absence A c sight surprised (S) 6 5 not surprised (S c ) Is being surprised independent of the form of the question asked? = P(A) = P(S) = P(S and A) = 6 91 = P(A) P(S) = P(S A) = 6/91 50/91 = =

83 Review More Penetrance of a Disease In a simple model of a recessive disease, each person has one of three genotypes (AA, Aa, aa) based on which of two forms each of their two genes (one from each parent) has, those of type aa have the disease, the others did not. A more realistic model assigns to each of these types a probability of getting the disease. Definition (Penetrance) The following probabilities are said to describe the penetrance of the disease: P(D AA) P(D Aa) P(D aa)

84 Review More Penetrance of a Disease Example (Consider the following penetrance model) P(D AA) = 0.01 P(D Aa) = 0.05 P(D aa) = Now consider an AA aa cross. What is the probability that a child will have the disease?

85 Review More Penetrance of a Disease Example (Consider the following penetrance model) P(D AA) = 0.01 P(D Aa) = 0.05 P(D aa) = Now consider an AA aa cross. What is the probability that a child will have the disease? P(D) = P(D Aa) = 0.05 because child will be Aa

86 Review More Penetrance of a Disease Example (Consider the following penetrance model) P(D AA) = 0.01 P(D Aa) = 0.05 P(D aa) = Now consider an AA aa cross. What is the probability that a child will have the disease? 2 In an AA Aa cross?

87 Review More Penetrance of a Disease Example (Consider the following penetrance model) P(D AA) = 0.01 P(D Aa) = 0.05 P(D aa) = Now consider an AA aa cross. What is the probability that a child will have the disease? 2 In an AA Aa cross? P(D) = P(D and AA) + P(D and Aa) + P(D and aa) = P(AA) P(D AA) + P(Aa) P(D Aa) + P(aa) P(D aa) = (.5)(.01) + (.5)(.05) + (0)(.5) =.03

88 Review More Penetrance of a Disease Example (Consider the following penetrance model) P(D AA) = 0.01 P(D Aa) = 0.05 P(D aa) = Now consider an AA aa cross. What is the probability that a child will have the disease? 2 In an AA Aa cross? 3 In an Aa Aa cross?

89 Review More Penetrance of a Disease Example (Consider the following penetrance model) P(D AA) = 0.01 P(D Aa) = 0.05 P(D aa) = Now consider an AA aa cross. What is the probability that a child will have the disease? 2 In an AA Aa cross? 3 In an Aa Aa cross? P(D) = P(D and AA) + P(D and Aa) + P(D and aa) = P(AA) P(D AA) + P(Aa) P(D Aa) + P(aa) P(D aa) = (.25)(.01) + (.5)(.05) + (.25)(.5) =.1525

90 Review More Genetics Example 2: DMD DMD is a serious form of Muscular Dystrophy, a sex-linked recessive disease. If a woman is a carrier, her sons have a 50% chance of being affected (daughters have a 50% chance of being carriers). 2/3 of DMD cases are inherited (from Mom); 1/3 of DMD cases are due to spontaneous mutations. Now suppose there is a screening test such that P(T + C+) =.7 [sensitivity how often correct when person is carrier] P(T C ) =.9 [specificity how often correct when person isn t carrier] If a woman has a child with DMD and then has a screening test done and it comes back positive, what is the probability that she is a carrier?

91 Review More Genetics Example 2: DMD Goal: Compute P(C + T +) = Data: P(C+and T +) P(T +) P(C+) = 2/3 [based on prior information] P(T + C+) =.7 [sensitivity] P(T C ) =.9 [specificity]

92 Review More Genetics Example 2: DMD Goal: Compute P(C + T +) = Data: Results: P(C+and T +) P(T +) P(C+) = 2/3 [based on prior information] P(T + C+) =.7 [sensitivity] P(T C ) =.9 [specificity] P(T +and C+) = P(C+) P(T + C+) = (2/3)(0.7) = 14/30 P(T + and C ) = P(C ) P(T + C ) = (1/3)(0.1) = 1/30 So P(C + T +) = 14/30 15/30 = 14/15 =

93 Review More Genetics Example 2: DMD Goal: Compute P(C + T +) = Data: Results: P(C+and T +) P(T +) P(C+) = 2/3 [based on prior information] P(T + C+) =.7 [sensitivity] P(T C ) =.9 [specificity] P(T +and C+) = P(C+) P(T + C+) = (2/3)(0.7) = 14/30 P(T + and C ) = P(C ) P(T + C ) = (1/3)(0.1) = 1/30 So P(C + T +) = 14/30 15/30 = 14/15 = What if the screening test is negative?

94 Review More Genetics Example 2: DMD Goal: Compute P(C + T +) = Data: Results: P(C+and T +) P(T +) P(C+) = 2/3 [based on prior information] P(T + C+) =.7 [sensitivity] P(T C ) =.9 [specificity] P(T +and C+) = P(C+) P(T + C+) = (2/3)(0.7) = 14/30 P(T + and C ) = P(C ) P(T + C ) = (1/3)(0.1) = 1/30 So P(C + T +) = 14/30 15/30 = 14/15 = What if the screening test is negative? Compute P(C + T ) instead. Same methods work. (A. 6/15)

95 Review More Keeping Organized: Tables You can keep your work organized on a problem like this using a table. C+ C total T + T total Fill in the boxes on the table as you figure them out, keeping track of the ones you need to know. In this case we want to know P(C + T +), so we need the values in the T + column.

96 Review More Keeping Organized: Tables You can keep your work organized on a problem like this using a table. T + T total C+ C total 1 Fill in the boxes on the table as you figure them out, keeping track of the ones you need to know. In this case we want to know P(C + T +), so we need the values in the T + column.

97 Review More Keeping Organized: Tables You can keep your work organized on a problem like this using a table. T + T total C+ 2/3 C 1/3 total 1 P(C+) = 2 3, P(C ) = 1 3 Fill in the boxes on the table as you figure them out, keeping track of the ones you need to know. In this case we want to know P(C + T +), so we need the values in the T + column.

98 Review More Keeping Organized: Tables You can keep your work organized on a problem like this using a table. T + T total C+ 14/30 2/3 C 1/30 1/3 total 1 P(C+) = 2 3, P(C ) = 1 3 Fill in the boxes on the table as you figure them out, keeping track of the ones you need to know. In this case we want to know P(C + T +), so we need the values in the T + column. P(T +and C+) = P(C+) P(T + C+) = (2/3)(0.7) = 14/30 P(T + and C ) = P(C ) P(T + C ) = (1/3)(0.1) = 1/30

99 Review More Keeping Organized: Tables You can keep your work organized on a problem like this using a table. T + T total C+ 14/30 2/3 C 1/30 1/3 total 15/30 1 P(C+) = 2 3, P(C ) = 1 3 Fill in the boxes on the table as you figure them out, keeping track of the ones you need to know. In this case we want to know P(C + T +), so we need the values in the T + column. P(T +and C+) = P(C+) P(T + C+) = (2/3)(0.7) = 14/30 P(T + and C ) = P(C ) P(T + C ) = (1/3)(0.1) = 1/30 P(T +) = P(T + and C+) + P(T + and C ) = 15/30

100 Review More Keeping Organized: Tables You can keep your work organized on a problem like this using a table. T + T total C+ 14/30 2/3 C 1/30 1/3 total 15/30 1 P(C+) = 2 3, P(C ) = 1 3 Fill in the boxes on the table as you figure them out, keeping track of the ones you need to know. In this case we want to know P(C + T +), so we need the values in the T + column. P(T +and C+) = P(C+) P(T + C+) = (2/3)(0.7) = 14/30 P(T + and C ) = P(C ) P(T + C ) = (1/3)(0.1) = 1/30 P(T +) = P(T + and C+) + P(T + and C ) = 15/30 So P(C + T +) = 14/30 15/30 = 14/15 =

101 Review More Keeping Organized: Trees Alternatively, we can represent this information in a tree.

102 Review More Generalized Product Rule Generalized Product Rule P(A and B and C) = P(A) P(B A) P(C A and B) }{{} =P(A and B) Example (Red Cards) A standard deck of cards has 52 cards (26 red; 26 black). If you are dealt 5 cards at random, what is the probability that all 5 are red?

103 Review More Generalized Product Rule Generalized Product Rule P(A and B and C) = P(A) P(B A) P(C A and B) }{{} =P(A and B) Example (Red Cards) A standard deck of cards has 52 cards (26 red; 26 black). If you are dealt 5 cards at random, what is the probability that all 5 are red? P(1st red) P(2nd red 1st red) P(3nd red 1st and 2nd red) P(4th red 1st, 2nd and 3rd are red) P(5th red 1st, 2nd, 3rd and 4th are red) = = < (1/2)5 48

104 Inclusion-Exclusion Review More Inclusion-Exclusion Rule P(A or B or C) = P(A) + P(B) + P(C) Things in none of the events never get counted. Things in one event get counted once. Things in two events get counted twice Things in all three events get counted 3 times

105 Inclusion-Exclusion Review More Inclusion-Exclusion Rule P(A or B or C) = P(A) + P(B) + P(C) P(A and B) P(A and C) P(B and C) Things in none of the events never get counted. Things in one event get counted once. Things in two events get counted twice, then uncounted once. Things in all three events get counted 3 times, then uncounted 3 times

106 Inclusion-Exclusion Review More Inclusion-Exclusion Rule P(A or B or C) = P(A) + P(B) + P(C) P(A and B) P(A and C) P(B and C) + P(A and B and C) Things in none of the events never get counted. Things in one event get counted once. Things in two events get counted twice, then uncounted once. Things in all three events get counted 3 times, then uncounted 3 times, then counted 1 time, for a net of one time counted.